2024 Edexcel IGCSE Mathematics (Specification A) Practice Paper with Answers

To simplify the fraction, we factorise both the numerator and the denominator. The numerator is a quadratic: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). The denominator can first be factorised by taking out the common factor of 2: \(2x^2 - 18 = 2(x^2 - 9)\). We then recognise the difference of two squares in the bracket: \(x^2 - 9 = (x - 3)(x + 3)\), which gives the fully factorised denominator as \(2(x - 3)(x + 3)\). Writing the fraction with these factorised forms, we get \(\frac{(2x + 1)(x - 3)}{2(x - 3)(x + 3)}\). Cancelling the common factor of \((x - 3)\) from the numerator and denominator gives \(\frac{2x + 1}{2(x + 3)}\), which can also be written as \(\frac{2x + 1}{2x + 6}\).

M1 for factorising the numerator as \((2x + 1)(x - 3)\) (or equivalent). M1 for factorising the denominator by taking out 2, i.e., \(2(x^2 - 9)\). M1 for fully factorising the denominator as \(2(x - 3)(x + 3)\). A1 for the correct simplified fraction \(\frac{2x+1}{2x+6}\) or \(\frac{2x+1}{2(x+3)}\).

First, we find the length of the diagonal of the base, \(AC\), using Pythagoras' Theorem in the right-angled triangle \(ABC\): \(AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100\), so \(AC = \sqrt{100} = 10\text{ cm}\). Next, we consider the right-angled triangle \(ACG\), where the angle \(\angle GAC\) is the angle that the diagonal \(AG\) makes with the base. Using trigonometry: \(\tan(\angle GAC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CG}{AC} = \frac{12}{10} = 1.2\). Therefore, \(\angle GAC = \tan^{-1}(1.2) = 50.1944...^\circ\). Correct to 1 decimal place, the angle is \(50.2^\circ\).

M1 for using Pythagoras' Theorem to calculate the base diagonal length \(AC^2 = 8^2 + 6^2\). A1 for obtaining \(AC = 10\). M1 for setting up a correct trigonometric ratio, e.g., \(\tan(\theta) = \frac{12}{10}\). A1 for a correct final answer of 50.2 (accept answers in range 50.1 to 50.3).

Let the original amount invested be \(P\). With compound interest of \(2.5\%\) per year, the value of the investment after 3 years is given by \(P \times 1.025^3\). We set up the equation: \(P \times 1.025^3 = 8615.12\). Rearranging to solve for \(P\), we get: \(P = \frac{8615.12}{1.025^3} = \frac{8615.12}{1.076890625} = 8000.0016...\). Rounding to the nearest penny gives \(\pounds 8000.00\).

M1 for identifying the multiplier \(1.025\). M1 for setting up the equation \(P \times 1.025^3 = 8615.12\). M1 for rearranging the equation to solve for \(P\), e.g., \(P = \frac{8615.12}{1.025^3}\). A1 for the correct final answer of 8000.00 (or 8000).

We use the Cosine Rule to find the length of \(PR\): \(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR)\). Substituting the given values: \(PR^2 = 7.4^2 + 11.2^2 - 2 \times 7.4 \times 11.2 \times \cos(115^\circ)\). This simplifies to: \(PR^2 = 54.76 + 125.44 - 165.76 \times (-0.422618...)\) which gives \(PR^2 = 180.20 + 70.053... = 250.253...\). Taking the square root: \(PR = \sqrt{250.253...} = 15.819...\). Correct to 3 significant figures, the length is \(15.8\text{ cm}\).

M1 for substituting the correct values into the Cosine Rule: \(7.4^2 + 11.2^2 - 2 \times 7.4 \times 11.2 \times \cos(115)\). A1 for evaluating intermediate steps to obtain \(PR^2 \approx 250.25\). M1 for taking the square root of their value of \(PR^2\). A1 for a correct final answer of 15.8 (accept answers in range 15.8 to 15.82).

Let the original price of the coat be \(X\). A reduction of \(20\%\) is represented by multiplying by \(0.80\). A further reduction of \(15\%\) on this sale price is represented by multiplying by \(0.85\). Therefore, the overall multiplier is \(0.80 \times 0.85 = 0.68\). We can set up the equation: \(0.68 \times X = 102\). Solving for \(X\): \(X = \frac{102}{0.68} = 150\). Thus, the original price was \(\pounds 150\).

M1 for writing a multiplier of \(0.80\) (or \(80\%\)) or \(0.85\) (or \(85\%\)). M1 for the combined multiplier \(0.80 \times 0.85 = 0.68\) (or equivalent method such as working backwards in two separate steps, e.g., \(102 \div 0.85 = 120\)). M1 for setting up the equation \(0.68X = 102\) or dividing \(120 \div 0.8\). A1 for the correct final answer of 150.

To make \(x\) the subject, we first multiply both sides by \((2 - x)\) to clear the fraction: \(y(2 - x) = 4x + 3\). Expanding the bracket gives: \(2y - xy = 4x + 3\). Next, we gather all terms containing \(x\) on one side and other terms on the other side: \(2y - 3 = 4x + xy\). Factoring out \(x\) on the right side: \(2y - 3 = x(4 + y)\). Finally, we divide by \((4 + y)\) to isolate \(x\): \(x = \frac{2y - 3}{4 + y}\).

M1 for multiplying both sides by \((2 - x)\) to get \(y(2 - x) = 4x + 3\). M1 for expanding the bracket correctly to get \(2y - xy = 4x + 3\). M1 for isolating terms with \(x\) on one side and factorising to get \(x(4 + y) = 2y - 3\) (or equivalent). A1 for the correct final formula \(x = \frac{2y - 3}{4 + y}\) or equivalent.

The area of a rectangle is length times width, so: \((2x + 5)(x - 1) = 30\). Expanding the brackets: \(2x^2 - 2x + 5x - 5 = 30\), which simplifies to \(2x^2 + 3x - 5 = 30\). Subtracting 30 from both sides gives the quadratic equation: \(2x^2 + 3x - 35 = 0\). We can solve this by factorising. We need two numbers that multiply to \(2 \times (-35) = -70\) and add to 3. These numbers are 10 and -7. Thus, \(2x^2 + 10x - 7x - 35 = 0 \implies 2x(x + 5) - 7(x + 5) = 0 \implies (2x - 7)(x + 5) = 0\). This gives two possible solutions: \(2x - 7 = 0 \implies x = 3.5\) or \(x + 5 = 0 \implies x = -5\). Since the width of a rectangle must be positive, \(x - 1 = -5 - 1 = -6\) is impossible, so we discard \(x = -5\). Therefore, \(x = 3.5\).

M1 for setting up the equation \((2x + 5)(x - 1) = 30\). M1 for expanding brackets and rearranging to form the quadratic equation \(2x^2 + 3x - 35 = 0\). M1 for solving the quadratic equation to get \(x = 3.5\) and \(x = -5\) (by factorisation, formula, or completing the square). A1 for identifying \(x = 3.5\) as the only valid solution.

We start by finding the expression for the composite function \(fg(x)\), which means substituting \(g(x)\) into \(f(x)\): \(fg(x) = f(g(x)) = f(2x + 5) = \frac{3}{(2x + 5) - 2} = \frac{3}{2x + 3}\). Next, we set this expression equal to \(\frac{1}{3}\): \(\frac{3}{2x + 3} = \frac{1}{3}\). Multiplying both sides by \(3(2x + 3)\) to clear the denominators gives: \(3 \times 3 = 1 \times (2x + 3) \implies 9 = 2x + 3\). Subtracting 3 from both sides gives \(2x = 6\), and dividing by 2 gives \(x = 3\).

M1 for attempting to find the composite function \(fg(x)\) by substituting \(g(x)\) into \(f(x)\). A1 for simplifying the composite function to \(\frac{3}{2x + 3}\). M1 for setting up the equation \(\frac{3}{2x + 3} = \frac{1}{3}\) and converting it to a linear equation, e.g., \(2x + 3 = 9\). A1 for the correct final answer of 3.

First, factorise each of the quadratic expressions:

1) \(6x^2 - 7x - 3 = (3x + 1)(2x - 3)\)

2) \(9x^2 - 1 = (3x - 1)(3x + 1)\) (difference of two squares)

3) \(2x^2 - 5x + 3 = (2x - 3)(x - 1)\)

4) \(x^2 - 2x + 1 = (x - 1)^2\)

Substitute these back into the original expression:

$$\frac{(3x + 1)(2x - 3)}{(3x - 1)(3x + 1)} \div \frac{(2x - 3)(x - 1)}{(x - 1)^2}$$

Simplify the individual fractions:

$$\frac{2x - 3}{3x - 1} \div \frac{2x - 3}{x - 1}$$

To divide by a fraction, multiply by its reciprocal:

$$\frac{2x - 3}{3x - 1} \times \frac{x - 1}{2x - 3}$$

Cancel out the common term \(2x - 3\):

$$\frac{x - 1}{3x - 1}$$

M1 for factorising \(6x^2 - 7x - 3\) correctly as \((3x + 1)(2x - 3)\) or \(9x^2 - 1\) as \((3x - 1)(3x + 1)\)

M1 for factorising \(2x^2 - 5x + 3\) correctly as \((2x - 3)(x - 1)\) or \(x^2 - 2x + 1\) as \((x - 1)^2\)

M1 for multiplying by the reciprocal and showing intention to cancel common terms

A1 for the fully simplified fraction \(\frac{x-1}{3x-1}\)

First, find the length of the diagonal \(AC\) on the horizontal base \(ABCD\) using Pythagoras' Theorem:

\(AC^2 = AB^2 + BC^2\)

\(AC^2 = 6^2 + 8^2 = 36 + 64 = 100\)

\(AC = \sqrt{100} = 10\text{ cm}\)

The angle that the line \(AG\) makes with the plane \(ABCD\) is the angle \(GAC\) in the right-angled triangle \(ACG\).

In triangle \(ACG\), \(CG = 12\text{ cm}\) (opposite side) and \(AC = 10\text{ cm}\) (adjacent side).

\(\tan(\angle GAC) = \frac{CG}{AC} = \frac{12}{10} = 1.2\)

\(\angle GAC = \tan^{-1}(1.2) \approx 50.1944...^\circ\)

To 1 decimal place, the angle is \(50.2^\circ\).

M1 for calculating the length of the diagonal \(AC\) as \(10\text{ cm}\) (or showing \(AC = \sqrt{6^2 + 8^2}\))

M1 for setting up a correct trigonometric ratio to find the angle, e.g., \(\tan(\theta) = \frac{12}{10}\) or \(\sin(\theta) = \frac{12}{AG}\) with a calculation of \(AG = \sqrt{244}\)

M1 for attempting to find the angle using inverse trigonometric functions, e.g., \(\theta = \tan^{-1}(1.2)\)

A1 for \(50.2\) (accept answers in the range \(50.19\) to \(50.2\))

Let the original price of the laptop be \(P\).

After a 15% reduction, the price is:

\(P \times (1 - 0.15) = 0.85P\)

After a further 10% reduction, the price becomes:

\(0.85P \times (1 - 0.10) = 0.85P \times 0.90 = 0.765P\)

We are given that this final price is £535.50:

\(0.765P = 535.50\)

Solve for \(P\):

\(P = \frac{535.50}{0.765} = 700\)

Therefore, the normal price was £700.

M1 for expressing the price after the first reduction, e.g., \(0.85P\)

M1 for a method to calculate the combined multiplier, e.g., \(0.85 \times 0.9 = 0.765\)

M1 for a complete method to find \(P\), e.g., \(535.50 \div 0.765\)

A1 for \(700\)

Multiply both sides by \((5 + x^2)\):

\(y(5 + x^2) = 2 - 3x^2\)

Expand the brackets:

\(5y + yx^2 = 2 - 3x^2\)

Rearrange the equation to group all terms containing \(x^2\) on one side:

\(yx^2 + 3x^2 = 2 - 5y\)

Factorise \(x^2\) from the left-hand side:

\(x^2(y + 3) = 2 - 5y\)

Divide both sides by \((y + 3)\):

\(x^2 = \frac{2 - 5y}{y + 3}\)

Since \(x > 0\), take the positive square root of both sides:

\(x = \sqrt{\frac{2 - 5y}{y + 3}}\)

M1 for multiplying both sides by \((5 + x^2)\) to clear the fraction

M1 for expanding brackets and rearranging to get all terms in \(x^2\) on one side

M1 for factorising out \(x^2\) to get \(x^2(y + 3) = 2 - 5y\)

A1 for \(x = \sqrt{\frac{2 - 5y}{y + 3}}\) (or equivalent)

Use the Cosine Rule to find the length of the third side \(BC\):

\(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\)

\(BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(64^\circ)\)

\(BC^2 = 49 + 81 - 126 \times \cos(64^\circ)\)

\(BC^2 = 130 - 126 \times 0.438371...\)

\(BC^2 = 130 - 55.2348... = 74.7651...\)

\(BC = \sqrt{74.7651...} \approx 8.6467\text{ cm}\)

Now, calculate the perimeter of triangle \(ABC\):

\(\text{Perimeter} = AB + AC + BC\)

\(\text{Perimeter} = 7 + 9 + 8.6467 \approx 24.6467\text{ cm}\)

To 3 significant figures, this is \(24.6\text{ cm\}.

M1 for substituting the given values correctly into the Cosine Rule: \(BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(64^\circ)\)

M1 for correct evaluation of \(BC^2\) (approx. \(74.8\)) or \(BC\) (approx. \(8.65\))

M1 for adding the three side lengths together: \(7 + 9 + \text{their } BC\)

A1 for \(24.6\) (accept answers in the range \(24.6\) to \(24.7\))

First, calculate the number of female employees in Company A:

\(150 \times 0.60 = 90\) female employees

Next, calculate the number of female employees in Company B:

\(250 \times 0.48 = 120\) female employees

Find the total number of female employees in the merged company:

\(90 + 120 = 210\) female employees

Find the total number of employees in the merged company:

\(150 + 250 = 400\) total employees

Calculate the percentage of female employees in the merged company:

$$\frac{210}{400} \times 100 = 52.5\%$$

M1 for finding the number of females in Company A: \(150 \times 0.60 = 90\)

M1 for finding the number of females in Company B: \(250 \times 0.48 = 120\)

M1 for a complete method to find the percentage: \(\frac{\text{their } 90 + \text{their } 120}{150 + 250} \times 100\)

A1 for \(52.5\) (or \(52.5\%\))

Multiply all terms by the common denominator \((x + 3)(2x - 1)\) to eliminate the fractions:

\(4(2x - 1) + 3(x + 3) = (x + 3)(2x - 1)\)

Expand both sides of the equation:

LHS: \(8x - 4 + 3x + 9 = 11x + 5\)

RHS: \(2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\)

Set the two expanded expressions equal to each other:

\(2x^2 + 5x - 3 = 11x + 5\)

Rearrange into a standard quadratic equation of the form \(ax^2 + bx + c = 0\):

\(2x^2 + 5x - 11x - 3 - 5 = 0\)

\(2x^2 - 6x - 8 = 0\)

Divide the entire equation by 2:

\(x^2 - 3x - 4 = 0\)

Factorise the quadratic:

\((x - 4)(x + 1) = 0\)

Thus, the solutions are:

\(x = 4\) or \(x = -1\)

M1 for attempting to clear fractions by multiplying through by the common denominator

M1 for correctly expanding both sides of the equation to obtain \(11x + 5\) and \(2x^2 + 5x - 3\)

M1 for rearranging and simplifying to form a three-term quadratic equation equivalent to \(2x^2 - 6x - 8 = 0\) or \(x^2 - 3x - 4 = 0\)

A1 for both solutions: \(x = 4\) and \(x = -1\)

Let's find the interior angle \(ABC\) of triangle \(ABC\).

The bearing of B from A is \(075^\circ\). A line pointing directly South from B makes an angle of \(75^\circ\) with the line segment \(BA\) (alternate angles with the North-line at A).

The bearing of C from B is \(150^\circ\), which is measured clockwise from the North-line at B. Since South is \(180^\circ\), the line \(BC\) makes an angle of \(180^\circ - 150^\circ = 30^\circ\) with the South-line.

Alternatively, co-interior angles between the North-line at A and the North-line at B add up to \(180^\circ\). Since the bearing is \(075^\circ\), the angle from BA to the North-line at B (anti-clockwise) is \(180^\circ - 75^\circ = 105^\circ\).

The bearing of C from B is \(150^\circ\) clockwise from North.

The direction of BA is \(255^\circ\) clockwise from North.

Thus, the interior angle \(\angle ABC = 255^\circ - 150^\circ = 105^\circ\).

Now, use the Cosine Rule to find \(AC\):

\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)\)

\(AC^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(105^\circ)\)

\(AC^2 = 324 + 625 - 900 \times (-0.258819...)\)

\(AC^2 = 949 + 232.937... = 1181.937...\)

\(AC = \sqrt{1181.937...} \approx 34.379\text{ km}\)

To 3 significant figures, the distance is \(34.4\text{ km}\).

M1 for finding the correct interior angle \(\angle ABC = 105^\circ\) (can be shown on a diagram)

M1 for substituting their values into the Cosine Rule: \(AC^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(105^\circ)\)

M1 for correctly evaluating \(AC^2\) (approx. \(1181.9\)) or \(AC = \sqrt{1181.9}\)

A1 for \(34.4\) (accept answers in the range \(34.3\) to \(34.4\))

Let \(N\) be the normal price of the computer.
Since the price is reduced by 20% to £500:
\(0.80 \times N = 500\)
\(N = \frac{500}{0.80} = 625\)

This normal price of £625 is deposited into a bank account with compound interest rate of \(r\%\) per annum.
After 2 years, the value is £676:
\(625 \times \left(1 + \frac{r}{100}\right)^2 = 676\)

Divide both sides by 625:
\(\left(1 + \frac{r}{100}\right)^2 = \frac{676}{625}\)

Take the square root of both sides:
\(1 + \frac{r}{100} = \sqrt{1.0816}\)
\(1 + \frac{r}{100} = 1.04\)

Subtract 1 from both sides:
\(\frac{r}{100} = 0.04\)
\(r = 4\)

M1 for a method to find the normal price, e.g., \(500 / 0.8\) (to get 625)
M1 for setting up the compound interest equation, e.g., \(625 \times (1 + r/100)^2 = 676\)
M1 for a method to solve for \(1 + r/100\), e.g., \(\sqrt{676/625}\)
A1 for \(r = 4\)

Multiply both sides by \((2 - ek)\):
\(H(2 - ek) = 4k - 3\)

Expand the brackets:
\(2H - eHk = 4k - 3\)

Rearrange to get all terms with \(k\) on one side:
\(2H + 3 = 4k + eHk\)

Factorise \(k\) from the right-hand side:
\(2H + 3 = k(4 + eH)\)

Divide by \((4 + eH)\) to make \(k\) the subject:
\(k = \frac{2H + 3}{4 + eH}\)

M1 for multiplying both sides by the denominator, e.g., \(H(2 - ek) = 4k - 3\)
M1 for expanding brackets correctly, e.g., \(2H - eHk = 4k - 3\)
M1 for collecting terms in \(k\) on one side and factorising \(k\), e.g., \(k(4 + eH) = 2H + 3\)
A1 for \(k = \frac{2H + 3}{4 + eH}\) (or equivalent, e.g., \(k = \frac{-2H - 3}{-4 - eH}\))

In right-angled triangle \(ABD\):
\(\tan(38^\circ) = \frac{BD}{AD}\)
\(BD = 8.4 \times \tan(38^\circ)\)
\(BD \approx 8.4 \times 0.7812856 = 6.5628\text{ m}\)

In right-angled triangle \(BCD\), \(BD\) is the hypotenuse:
\(\sin(47^\circ) = \frac{CD}{BD}\)
\(CD = BD \times \sin(47^\circ)\)
\(CD \approx 6.5628 \times \sin(47^\circ)\)
\(CD \approx 6.5628 \times 0.7313537 = 4.8001\text{ m}\)

To 3 significant figures, the length of \(CD\) is \(4.80\text{ m}\).

M1 for using tangent ratio to find \(BD\), e.g., \(\tan(38^\circ) = \frac{BD}{8.4}\)
A1 for \(BD = 6.56(28...)\)
M1 for using sine ratio with their \(BD\), e.g., \(CD = BD \times \sin(47^\circ)\)
A1 for \(4.80\) (or \(4.8\))

Let \(n\) be the number of years.
The value of the machine after \(n\) years is given by:
\(V = 24000 \times (1 - 0.12)^n = 24000 \times (0.88)^n\)

We want to find the smallest integer \(n\) such that:
\(24000 \times (0.88)^n < 10000\)

Let's calculate the value for successive years:
For \(n = 5\):
\(V = 24000 \times (0.88)^5 \approx 12665.61\)

For \(n = 6\):
\(V = 24000 \times (0.88)^6 \approx 11145.74\)

For \(n = 7\):
\(V = 24000 \times (0.88)^7 \approx 9808.25\)

Since 9808.25 is less than 10 000, it takes 7 years.

M1 for a correct multiplier of \(0.88\) or an expression like \(24000 \times (0.88)^n\)
M1 for a method of trial to find the value at \(n \ge 5\), showing at least one calculation, e.g., \(24000 \times 0.88^5\)
M1 for calculating the values for \(n = 6\) and \(n = 7\)
A1 for 7

Factorise the numerator \(3x^2 - 10x - 8\):
We look for two numbers that multiply to \(3 \times (-8) = -24\) and add to \(-10\). These are \(-12\) and \(2\).
\(3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)\)

Factorise the denominator \(2x^2 - 7x - 4\):
We look for two numbers that multiply to \(2 \times (-4) = -8\) and add to \(-7\). These are \(-8\) and \(1\).
\(2x^2 - 8x + x - 4 = 2x(x - 4) + 1(x - 4) = (2x + 1)(x - 4)\)

Write the fraction with the factorised terms:
\(\frac{(3x + 2)(x - 4)}{(2x + 1)(x - 4)}\)

Cancel the common factor \((x - 4)\):
\(\frac{3x + 2}{2x + 1}\)

M2 for factorising the numerator correctly as \((3x + 2)(x - 4)\) (M1 for \((3x+a)(x+b)\) where \(ab=-8\) or \(3b+a=-10\))
M1 for factorising the denominator correctly as \((2x + 1)(x - 4)\) (or M1 for \((2x+c)(x+d)\) where \(cd=-4\) or \(2d+c=-7\))
A1 for \(\frac{3x + 2}{2x + 1}\)

First, use the Cosine Rule to find the angle \(B\) (angle \(ABC\)):
\(b^2 = a^2 + c^2 - 2ac \cos(B)\)
Here, \(b = AC = 11.4\), \(a = BC = 8.5\), and \(c = AB = 7.2\).

\(11.4^2 = 8.5^2 + 7.2^2 - 2 \times 8.5 \times 7.2 \times \cos(B)\)
\(129.96 = 72.25 + 51.84 - 122.4 \cos(B)\)
\(129.96 = 124.09 - 122.4 \cos(B)\)
\(122.4 \cos(B) = 124.09 - 129.96\)
\(122.4 \cos(B) = -5.87\)
\(\cos(B) = -\frac{5.87}{122.4} \approx -0.0479575\)

\(B = \arccos(-0.0479575) \approx 92.7484^\circ\)

Now, calculate the area of the triangle:
\(\text{Area} = \frac{1}{2} a c \sin(B) = \frac{1;}{2} \times 8.5 \times 7.2 \times \sin(92.7484^\circ)\)
\(\text{Area} \approx 30.6 \times 0.99885 = 30.5648\text{ cm}^2\)

Correct to 3 significant figures, the area is \(30.6\text{ cm}^2\).

M1 for a correct substitution into the Cosine Rule, e.g., \(11.4^2 = 8.5^2 + 7.2^2 - 2 \times 8.5 \times 7.2 \times \cos(B)\)
M1 for a method to calculate \(\cos(B)\), leading to \(\cos(B) = -0.048\) or \(B = 92.7^\circ\)
M1 for a method to calculate the area using \(\frac{1}{2} a c \sin(B)\)
A1 for \(30.6\)

The area of a triangle is given by:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

Substitute the given values:
\(\frac{1}{2}(2x - 3)(x + 4) = 22\)

Multiply both sides by 2:
\((2x - 3)(x + 4) = 44\)

Expand the brackets:
\(2x^2 + 8x - 3x - 12 = 44\)
\(2x^2 + 5x - 12 = 44\)

Subtract 44 from both sides to form a quadratic equation:
\(2x^2 + 5x - 56 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-56)}}{2(2)}\)
\(x = \frac{-5 \pm \sqrt{25 + 448}}{4} = \frac{-5 \pm \sqrt{473}}{4}\)

Since \(x\) must be positive for the base \(2x - 3\) to be positive:
\(x = \frac{-5 + 21.74856}{4} \approx 4.187\)

To 3 significant figures, \(x = 4.19\).

M1 for setting up the area equation, e.g., \(\frac{1}{2}(2x - 3)(x + 4) = 22\)
M1 for expanding and rearranging to form a 3-term quadratic equation, e.g., \(2x^2 + 5x - 56 = 0\)
M1 for a correct substitution into the quadratic formula
A1 for \(4.19\)

(a) To find \(fg(x)\), substitute \(g(x)\) into \(f(x)\):
\(fg(x) = f(3x + 2) = \frac{5}{(3x + 2) - 3} = \frac{5}{3x - 1}\)

(b) To solve \(fg(x) = x\):
\rac{5}{3x - 1} = x\)
\(5 = x(3x - 1)\)
\(5 = 3x^2 - x\)
\(3x^2 - x - 5 = 0\)

Use the quadratic formula:
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-5)}}{2(3)}\)
\(x = \frac{1 \pm \sqrt{1 + 60}}{6} = \frac{1 \pm \sqrt{61}}{6}\)

\(x_1 = \frac{1 + 7.81025}{6} \approx 1.47\)
\(x_2 = \frac{1 - 7.81025}{6} \approx -1.14\)

M1 for finding \(fg(x)\) as \(\frac{5}{(3x + 2) - 3}\)
A1 for simplifying \(fg(x)\) to \(\frac{5}{3x - 1}\)
M1 for setting up the equation \(\frac{5}{3x - 1} = x\) and forming the quadratic equation \(3x^2 - x - 5 = 0\)
A1 for \(x = 1.47\) and \(x = -1.14\)

First, factorise the denominators of both fractions:

1) \(x^2 - 4 = (x - 2)(x + 2)\) (difference of two squares)
2) \(x^2 - 2x = x(x - 2)\)

Now rewrite the subtraction with these factorised denominators:

$$\frac{2}{(x - 2)(x + 2)} - \frac{1}{x(x - 2)}$$

The lowest common denominator is \(x(x - 2)(x + 2)\). Express each fraction with this denominator:

$$\frac{2x}{x(x - 2)(x + 2)} - \frac{1(x + 2)}{x(x - 2)(x + 2)}$$

Combine the numerators:

$$\frac{2x - (x + 2)}{x(x - 2)(x + 2)}$$

$$\frac{2x - x - 2}{x(x - 2)(x + 2)} = \frac{x - 2}{x(x - 2)(x + 2)}$$

Cancel the common factor of \((x - 2)\) from both the numerator and the denominator:

$$\frac{1}{x(x + 2)}$$

This can also be written as:

$$\frac{1}{x^2 + 2x}$$

M1: For factorising at least one denominator correctly: \(x^2 - 4 = (x-2)(x+2)\) or \(x^2 - 2x = x(x-2)\)
M1: For writing both fractions with a common denominator, e.g., \(\frac{2x - (x+2)}{x(x-2)(x+2)}\) or \(\frac{2(x^2-2x) - (x^2-4)}{(x^2-4)(x^2-2x)}\)
A1: For simplifying the numerator to \(x-2\) to get \(\frac{x-2}{x(x-2)(x+2)}\)
A1: For the final simplified fraction: \(\frac{1}{x(x+2)}\) or \(\frac{1}{x^2+2x}\)

We use the compound interest formula: \(A = P\left(1 + \frac{r}{100}\right)^t\). Here, \(P = 6200\), \(A = 6822.41\), and \(t = 3\). Substituting these values: \(6200\left(1 + \frac{r}{100}\right)^3 = 6822.41\). Divide both sides by 6200: \(\left(1 + \frac{r}{100}\right)^3 = \frac{6822.41}{6200} = 1.1003887...\). Take the cube root of both sides: \(1 + \frac{r}{100} = \sqrt[3]{1.1003887...} \approx 1.0323999...\). Subtract 1 from both sides: \(\frac{r}{100} = 0.0323999...\). Multiply by 100: \(r \approx 3.24\%\).

M1: for setting up the correct equation \(6200 \times x^3 = 6822.41\) (where \(x = 1 + r/100\))
M1: for \(x^3 = 1.1003887...\)
M1: for \(x = 1.0323999...\)
A1: for 3.24

First, factorise the numerator \(3x^2 - 14x - 5\). We look for two numbers that multiply to \(3 \times (-5) = -15\) and add to \(-14\). These numbers are \(-15\) and \(1\). So, \(3x^2 - 15x + x - 5 = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5)\). Next, factorise the denominator \(9x^2 - 1\). This is a difference of two squares: \(9x^2 - 1 = (3x - 1)(3x + 1)\). Now, write the fraction with the factorised forms: \(\frac{(3x + 1)(x - 5)}{(3x - 1)(3x + 1)}\). Cancel the common factor of \((3x + 1)\) from both numerator and denominator: \(\frac{x - 5}{3x - 1}\).

M1: for factorising numerator into two brackets where one is \((3x + 1)\) or \((x - 5)\)
M1: for factorising denominator to \((3x - 1)(3x + 1)\)
M1: for showing the cancellation of the common factor \((3x + 1)\)
A1: for \(\frac{x - 5}{3x - 1}\)

First, find the interior angle \(ABC\). Let the North line at \(B\) be drawn. The angle between North and the line \(BA\) is \(48^{\circ}\) (alternate angles). The bearing of \(C\) from \(B\) is \(135^{\circ}\). The angle between the South line at \(B\) and \(BA\) is \(48^{\circ}\). The angle between the South line at \(B\) and \(BC\) is \(180^{\circ} - 135^{\circ} = 45^{\circ}\). Thus, the angle \(ABC\) is \(48^{\circ} + 45^{\circ} = 93^{\circ}\). Now, apply the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(ABC)\). Substituting the values: \(AC^2 = 5.2^2 + 7.8^2 - 2 \times 5.2 \times 7.8 \times \cos(93^{\circ})\). \(AC^2 = 27.04 + 60.84 - 81.12 \times (-0.052335956)\). \(AC^2 = 87.88 + 4.24549 = 92.12549\). \(AC = \sqrt{92.12549} \approx 9.60\text{ km}\).

M1: for calculating the correct angle \(ABC = 93^{\circ}\) (or angle shown on diagram)
M1: for substituting values correctly into the Cosine Rule: \(5.2^2 + 7.8^2 - 2(5.2)(7.8)\cos(93^{\circ})\)
M1: for evaluating the terms to find \(AC^2 \approx 92.1\)
A1: for 9.60 (accept answers in range 9.59 to 9.61)

Let \(P\) be the original price of the jacket. The price after a \(15\%\) reduction is \(0.85P\). The price after a further \(10\%\) reduction is \(0.85P \times 0.90 = 0.765P\). We are given that this final price is £107.10. Therefore, \(0.765P = 107.10\). To find \(P\), divide 107.10 by 0.765: \(P = \frac{107.10}{0.765} = 140\).

M1: for recognizing the loyalty card price is \(90\%\) of the sale price and calculating \(107.10 \div 0.90 = 119\) or writing \(P \times 0.85 \times 0.90 = 107.10\)
M1: for recognizing the sale price is \(85\%\) of the original price and writing \(119 \div 0.85\) or \(P = 107.10 \div 0.765\)
M1: for an evaluation step, e.g., finding the intermediate price of 119 or the combined multiplier of 0.765
A1: for 140

First, multiply both sides by \((2w + 5)\) to clear the fraction: \(y(2w + 5) = 4 - 3w\). Expand the bracket on the left side: \(2wy + 5y = 4 - 3w\). Next, move all terms containing \(w\) to one side, and all other terms to the opposite side: \(2wy + 3w = 4 - 5y\). Factorise \(w\) out of the left-hand side: \(w(2y + 3) = 4 - 5y\). Finally, divide both sides by \((2y + 3)\) to isolate \(w\): \(w = \frac{4 - 5y}{2y + 3}\).

M1: for multiplying by \(2w + 5\) to get \(y(2w + 5) = 4 - 3w\)
M1: for expanding and rearranging to collect terms in \(w\) on one side, e.g., \(2wy + 3w = 4 - 5y\)
M1: for factorising \(w\) to get \(w(2y + 3) = 4 - 5y\)
A1: for \(w = \frac{4 - 5y}{2y + 3}\) (or equivalent, e.g., \(w = \frac{5y - 4}{-2y - 3}\))

Let \(M\) be the center of the square base \(ABCD\). The diagonal of the base \(AC\) can be found using Pythagoras' theorem on triangle \(ADC\): \(AC^2 = 6^2 + 6^2 = 72 \implies AC = \sqrt{72} = 6\sqrt{2} \approx 8.485\text{ cm}\). The distance from corner \(A\) to the center \(M\) is half of the diagonal: \(AM = \frac{AC}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \approx 4.243\text{ cm}\). In the right-angled triangle \(VMA\), the perpendicular height \(VM = 8\text{ cm}\). The angle \(\theta\) between the edge \(VA\) and the base is angle \(VAM\): \(\tan(\theta) = \frac{VM}{AM} = \frac{8}{3\sqrt{2}}\). \(\tan(\theta) \approx 1.8856\). \(\theta = \tan^{-1}(1.8856) \approx 62.1^{\circ}\).

M1: for finding the diagonal of the base: \(AC = \sqrt{6^2 + 6^2} = \sqrt{72}\)
M1: for finding the half-diagonal: \(AM = \frac{\sqrt{72}}{2} = 3\sqrt{2} \approx 4.24\)
M1: for \(\tan(\theta) = \frac{8}{3\sqrt{2}}\) or equivalent trigonometric setup
A1: for 62.1 (accept answers in range 62.0 to 62.1)

The area of a rectangle is length multiplied by width: \((2x + 3)(x - 1) = 45\). Expand the brackets: \(2x^2 - 2x + 3x - 3 = 45 \implies 2x^2 + x - 3 = 45\). Rearrange into standard quadratic form \(ax^2 + bx + c = 0\): \(2x^2 + x - 48 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = 1\), and \(c = -48\): \(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-48)}}{2(2)}\). \(x = \frac{-1 \pm \sqrt{1 + 384}}{4} = \frac{-1 \pm \sqrt{385}}{4}\). Since the width \((x-1)\) must be positive, \(x\) must be greater than 1. Thus, we take the positive root: \(x = \frac{-1 + \sqrt{385}}{4} \approx \frac{-1 + 19.6214}{4} = 4.6553...\). To 3 significant figures, \(x = 4.66\).

M1: for setting up the equation \((2x + 3)(x - 1) = 45\)
M1: for expanding and rearranging to \(2x^2 + x - 48 = 0\)
M1: for substituting into the quadratic formula: \(x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-48)}}{4}\)
A1: for 4.66 (accept 4.65 to 4.67)

Let the total number of employees initially be \(T\). The number of female employees is \(0.60T\) and the number of male employees is \(0.40T\). After the changes: The new number of female employees is \(0.60T \times 1.15 = 0.69T\). The new number of male employees is \(0.40T \times 0.90 = 0.36T\). The new total number of employees is \(0.69T + 0.36T = 1.05T\). The percentage change in the total number of employees is \(\frac{1.05T - T}{T} \times 100 = 5\%\).

M1: for identifying that females represent \(60\%\) and males represent \(40\%\) (or using a numerical base like 100, so 60 females and 40 males)
M1: for calculating the new number of females \(0.60 \times 1.15 = 0.69\) (or 69) or the new number of males \(0.40 \times 0.90 = 0.36\) (or 36)
M1: for finding the new total: \(0.69 + 0.36 = 1.05\) (or 105)
A1: for 5 (or 5% increase)

First, multiply both sides by the denominator to clear the fraction: \(p(2t + a) = 5 - 3t\). Next, expand the brackets: \(2pt + ap = 5 - 3t\). Collect all terms containing \(t\) on one side of the equation and the other terms on the opposite side: \(2pt + 3t = 5 - ap\). Factorise \(t\) from the left side: \(t(2p + 3) = 5 - ap\). Finally, divide both sides by \(2p + 3\) to isolate \(t\): \(t = \frac{5 - ap}{2p + 3}\).

M1 for multiplying by \(2t + a\) to get \(p(2t + a) = 5 - 3t\). M1 for expanding brackets and grouping terms in \(t\) on one side. M1 for factorising \(t\) to get \(t(2p + 3) = 5 - ap\). A1 for the correct final formula \(t = \frac{5 - ap}{2p + 3}\) (or equivalent, e.g. \(t = \frac{ap - 5}{-2p - 3}\)).

First, find the length of the base diagonal \(AC\) using Pythagoras' theorem on the horizontal right-angled triangle \(ABC\): \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\). The diagonal \(EC\) makes an angle with the base \(ABCD\) which is the angle \(\angle ECA\). In the vertical right-angled triangle \(EAC\), the side opposite to this angle is \(AE = 8\text{ cm}\) and the adjacent side is \(AC = 13\text{ cm}\). Using trigonometry: \(\tan(\angle ECA) = \frac{AE}{AC} = \frac{8}{13}\). Thus, \(\angle ECA = \arctan\left(\frac{8}{13}\right) \approx 31.6075^\circ\). Correct to 1 decimal place, the angle is \(31.6^\circ\).

M1 for using Pythagoras to find the base diagonal: \(12^2 + 5^2\) or \(\sqrt{12^2 + 5^2}\). A1 for \(AC = 13\). M1 for \(\tan(\theta) = \frac{8}{13}\) (or using their \(AC\)). A1 for \(31.6\) (accept answers in the range 31.6 to 31.61).

The formula for compound interest is \(\text{Value} = P\left(1 + \frac{R}{100}\right)^n\). Substituting the given values: \(9261 = 8000\left(1 + \frac{R}{100}\right)^3\). Divide both sides by 8000: \(\left(1 + \frac{R}{100}\right)^3 = \frac{9261}{8000} = 1.157625\). Take the cube root of both sides: \(1 + \frac{R}{100} = \sqrt[3]{1.157625} = 1.05\). Subtract 1 from both sides: \(\frac{R}{100} = 0.05\). Multiply by 100 to find \(R\): \(R = 5\).

M1 for setting up the correct compound interest equation: \(8000(1 + R/100)^3 = 9261\). M1 for simplifying to \((1 + R/100)^3 = 1.157625\) (or \(\frac{9261}{8000}\)). M1 for taking the cube root: \(1 + R/100 = 1.05\). A1 for \(R = 5\).

Multiply all terms by the common denominator \((x+2)(x-1)\) to eliminate the fractions: \(5(x-1) + 3(x+2) = 2(x+2)(x-1)\). Expand both sides of the equation: \(5x - 5 + 3x + 6 = 2(x^2 + x - 2)\), which simplifies to \(8x + 1 = 2x^2 + 2x - 4\). Rearrange the equation into standard quadratic form \(ax^2 + bx + c = 0\): \(2x^2 - 6x - 5 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = -6\), and \(c = -5\): \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-5)}}{2(2)}\), which simplifies to \(x = \frac{6 \pm \sqrt{36 + 40}}{4} = \frac{6 \pm \sqrt{76}}{4}\). This gives two solutions: \(x = \frac{6 + 8.7178}{4} \approx 3.679\) and \(x = \frac{6 - 8.7178}{4} \approx -0.6794\). Correct to 3 significant figures, the solutions are \(3.68\) and \(-0.679\).

M1 for a correct equation without fractions: \(5(x-1) + 3(x+2) = 2(x+2)(x-1)\). M1 for expanding and simplifying to a standard quadratic equation: \(2x^2 - 6x - 5 = 0\). M1 for a correct substitution of their quadratic coefficients into the quadratic formula. A1 for both answers correct to 3 significant figures: \(3.68\) and \(-0.679\) (accept \(3.679\) and \(-0.6794\)).

First, use the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a c \sin(B)\). Substituting the given values: \(28 = \frac{1}{2} \times 10 \times 7 \times \sin(B)\), which simplifies to \(28 = 35 \sin(B)\). Therefore, \(\sin(B) = \frac{28}{35} = 0.8\). Since angle \(ABC\) is obtuse, the angle must be between \(90^\circ\) and \(180^\circ\): \(B = 180^\circ - \arcsin(0.8) \approx 180^\circ - 53.13^\circ = 126.87^\circ\). Now, use the cosine rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(B)\). Since \(\cos(126.87^\circ) = -0.6\), we have: \(AC^2 = 7^2 + 10^2 - 2(7)(10)(-0.6) = 49 + 100 + 84 = 233\). Thus, \(AC = \sqrt{233} \approx 15.264\text{ cm}\). Correct to 3 significant figures, \(AC = 15.3\text{ cm}\).

M1 for substituting area values into \(\frac{1}{2} a c \sin(B)\): \(28 = \frac{1}{2} \times 10 \times 7 \sin(B)\). A1 for finding the obtuse angle \(B \approx 126.87^\circ\) (or for stating \(\sin(B) = 0.8\) and \(\cos(B) = -0.6\)). M1 for substituting their values into the cosine rule: \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(126.87^\circ)\). A1 for \(15.3\) (accept answers in the range 15.2 to 15.3).

Let \(P\) represent the profit of the company in 2020. In 2021, the profit was \(1.15 \times P\). In 2022, the profit was \(0.92 \times (1.15 \times P)\), which simplifies to \(1.058 \times P\). We are given that the profit in 2022 was £243,340, so we set up the equation: \(1.058 \times P = 243340\). Solving for \(P\): \(P = \frac{243340}{1.058} = 230000\). Therefore, the profit in 2020 was £230,000.

M1 for writing an expression for the profit in 2021, e.g. \(1.15 \times P\). M1 for combining both percentage changes, e.g. \(1.15 \times 0.92 = 1.058\). M1 for setting up the equation \(1.058 \times P = 243340\) and dividing: \(\frac{243340}{1.058}\). A1 for \(230000\).

First, find the interior angle \(\angle BCD\) in triangle \(BCD\). The bearing of \(C\) from \(B\) is \(055^\circ\). The back bearing (the direction from \(C\) to \(B\)) is \(055^\circ + 180^\circ = 235^\circ\). Since the bearing of \(D\) from \(C\) is \(165^\circ\), the interior angle \(\angle BCD\) is the difference between these two directions: \(\angle BCD = 235^\circ - 165^\circ = 70^\circ\). Now, we can apply the cosine rule to triangle \(BCD\) to find the distance \(BD\): \(BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos(\angle BCD)\). Substitute \(BC = 8\), \(CD = 12\), and \(\angle BCD = 70^\circ\): \(BD^2 = 8^2 + 12^2 - 2(8)(12)\cos(70^\circ)\), \(BD^2 = 64 + 144 - 192\cos(70^\circ)\), \(BD^2 = 208 - 192(0.34202)\), \(BD^2 = 208 - 65.668 = 142.332\). Thus, \(BD = \sqrt{142.332} \approx 11.93\text{ km}\). Correct to 3 significant figures, the distance is \(11.9\text{ km}\).

M1 for finding the interior angle \(\angle BCD = 70^\circ\) (or showing equivalent working with parallel lines and bearings). M1 for a correct substitution into the Cosine Rule: \(BD^2 = 8^2 + 12^2 - 2 \times 8 \times 12 \cos(70^\circ)\). M1 for evaluating \(BD^2\) to at least 3 significant figures (e.g. 142 or 142.3). A1 for \(11.9\) (accept answers in the range 11.9 to 12.0).

First, factorise each quadratic expression: 1. \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). 2. \(2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3)\). 3. \(2x^2 + 7x + 3 = (2x + 1)(x + 3)\). 4. \(x^2 + 6x + 9 = (x + 3)^2\). Substitute these factorised forms back into the expression and change the division to multiplication by the reciprocal of the second fraction: \(\frac{(2x + 1)(x - 3)}{2(x - 3)(x + 3)} \times \frac{(x + 3)^2}{(2x + 1)(x + 3)}\). Combine the terms into a single fraction: \(\frac{(2x + 1)(x - 3)(x + 3)^2}{2(x - 3)(x + 3)(2x + 1)(x + 3)} = \frac{(2x + 1)(x - 3)(x + 3)^2}{2(2x + 1)(x - 3)(x + 3)^2}\). Cancel out all identical terms in the numerator and denominator: \((2x+1)\), \((x-3)\), and \((x+3)^2\) all cancel out, leaving \(\frac{1}{2}\).

M1 for factorising either \(2x^2 - 5x - 3\) to \((2x + 1)(x - 3)\) or \(2x^2 - 18\) to \(2(x - 3)(x + 3)\). M1 for factorising either \(2x^2 + 7x + 3\) to \((2x + 1)(x + 3)\) or \(x^2 + 6x + 9\) to \((x + 3)^2\). M1 for inverting the divisor and multiplying: \(\frac{(2x + 1)(x - 3)}{2(x - 3)(x + 3)} \times \frac{(x + 3)^2}{(2x + 1)(x + 3)}\). A1 for \(\frac{1}{2}\) (or \(0.5\)).

Let the amount in Account A be \(x\) and in Account B be \(x + 4000\). At the end of the year, the value in Account A is \(1.12x\) and the value in Account B is \(0.95(x + 4000)\). We are given that the sum of these values is £28,640. Setting up the equation: \(1.12x + 0.95(x + 4000) = 28640\). Expanding and simplifying: \(1.12x + 0.95x + 3800 = 28640\), which gives \(2.07x = 24840\). Solving for \(x\): \(x = 12000\).

M1: for expressing the end of year value of Account A as \(1.12x\) or Account B as \(0.95(x + 4000)\). M1: for setting up the correct equation \(1.12x + 0.95(x + 4000) = 28640\). M1: for simplifying to the form \(2.07x = 24840\). A1: for 12000 (accept £12,000).

Factorise the numerator: \(2x^2 + 5x - 12 = 2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\). Factorise the denominator: \(3x^2 - 48 = 3(x^2 - 16) = 3(x - 4)(x + 4)\). Divide out the common factor of \(x + 4\): \(\frac{(2x - 3)(x + 4)}{3(x - 4)(x + 4)} = \frac{2x - 3}{3(x - 4)}\).

M1: for factorising the numerator as \((2x - 3)(x + 4)\). M1: for factorising the denominator as \(3(x^2 - 16)\) or \(3(x - 4)(x + 4)\). M1: for writing the fully factorised fraction. A1: for \(\frac{2x - 3}{3(x - 4)}\) or \(\frac{2x - 3}{3x - 12}\).

First, calculate the length of the base diagonal \(BD\). Since \(ABCD\) is a rectangle with \(AB = 8\text{ cm}\) and \(AD = 6\text{ cm}\), using Pythagoras' theorem: \(BD = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\). The height \(DH\) is vertical, so triangle \(BDH\) is a right-angled triangle at \(D\). We are given that angle \(HBD = 40^\circ\). Using trigonometry: \(\tan(40^\circ) = \frac{DH}{BD} = \frac{DH}{10}\). Therefore, \(DH = 10 \times \tan(40^\circ) \approx 8.390996...\text{ cm}\). Correct to 3 significant figures, \(DH = 8.39\text{ cm}\).

M1: for a method to find the diagonal \(BD\), e.g., \(\sqrt{8^2 + 6^2}\). A1: for \(BD = 10\). M1: for using \(\tan(40^\circ) = \frac{DH}{10}\) or equivalent. A1: for 8.39 (accept answers in range 8.39 - 8.40).

Let the normal price of the coat be \(x\) pounds. The price after a 15% reduction is \(0.85x\). The price after an additional 10% reduction is \(0.85x \times 0.90 = 0.765x\). Setting up the equation: \(0.765x = 68.85\). Solving for \(x\): \(x = \frac{68.85}{0.765} = 90\). Thus, the normal price of the coat is £90.

M1: for multiplying 0.85 by 0.90 (to get 0.765) or for finding the price before card discount, i.e., \(68.85 \div 0.90 = 76.50\). M1: for setting up the equation \(0.765x = 68.85\) or \(0.85y = 76.50\). M1: for \(68.85 \div 0.765\) or \(76.50 \div 0.85\). A1: for 90.

First, multiply both sides by \((5w + 4)\) to clear the fraction: \(t(5w + 4) = 3 - 2w\). Expand the left side: \(5wt + 4t = 3 - 2w\). Move all terms involving \(w\) to one side: \(5wt + 2w = 3 - 4t\). Factorise \(w\) on the left side: \(w(5t + 2) = 3 - 4t\). Divide by \((5t + 2)\) to solve for \(w\): \(w = \frac{3 - 4t}{5t + 2}\).

M1: for multiplying both sides by the denominator, e.g., \(t(5w + 4) = 3 - 2w\). M1: for expanding and collecting terms with \(w\) on one side, e.g., \(5wt + 2w = 3 - 4t\). M1: for factorising \(w\), e.g., \(w(5t + 2) = 3 - 4t\). A1: for \(w = \frac{3 - 4t}{5t + 2}\) (or equivalent, e.g., \(w = \frac{4t - 3}{-5t - 2}\)).

Using the area formula: \(\text{Area} = \frac{1}{2} a c \sin(B)\), we have \(35 = \frac{1}{2} \times 12 \times 7 \times \sin(ABC)\). This simplifies to \(35 = 42 \sin(ABC)\), so \(\sin(ABC) = \frac{35}{42} = \frac{5}{6}\). Since the angle \(ABC\) is obtuse, \(ABC = 180^\circ - \arcsin(5/6) \approx 180^\circ - 56.443^\circ = 123.557^\circ\). Now, use the Cosine Rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the values: \(AC^2 = 7^2 + 12^2 - 2 \times 7 \times 12 \times \cos(123.557^\circ) = 49 + 144 - 168 \times (-0.55277) = 193 + 92.865 = 285.865\). Taking the square root: \(AC = \sqrt{285.865} \approx 16.9075...\text{ cm}\). Correct to 3 significant figures, \(AC = 16.9\text{ cm}\).

M1: for setting up the area equation \(\frac{1}{2} \times 12 \times 7 \times \sin(B) = 35\). M1: for finding the obtuse angle \(ABC \approx 123.6^\circ\) or for obtaining \(\cos(ABC) = -\frac{\sqrt{11}}{6}\). M1: for substituting values into the Cosine Rule, e.g., \(AC^2 = 7^2 + 12^2 - 2(7)(12)\cos(123.6^\circ)\). A1: for 16.9 (accept answers in range 16.9 - 17.0).

Multiply each term by the common denominator \((x+2)(x-1)\): \(5(x-1) + 3(x+2) = 2(x+2)(x-1)\). Expand the brackets: \(5x - 5 + 3x + 6 = 2(x^2 + x - 2)\), which simplifies to \(8x + 1 = 2x^2 + 2x - 4\). Rearrange into standard quadratic form: \(2x^2 - 6x - 5 = 0\). Solve using the quadratic formula: \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-5)}}{2(2)} = \frac{6 \pm \sqrt{36 + 40}}{4} = \frac{6 \pm \sqrt{76}}{4}\). This gives \(x = \frac{6 + 8.7178}{4} \approx 3.679\) and \(x = \frac{6 - 8.7178}{4} \approx -0.679\).

M1: for writing the LHS over a single common denominator or multiplying all terms by \((x+2)(x-1)\). M1: for expanding and simplifying to a quadratic equation, e.g., \(8x + 1 = 2(x^2 + x - 2)\). M1: for rearranging to a correct quadratic equation, e.g., \(2x^2 - 6x - 5 = 0\). A1: for both 3.68 and -0.679.

The bearing of \(B\) from \(A\) is \(060^\circ\). This means the back bearing of \(A\) from \(B\) is \(060^\circ + 180^\circ = 240^\circ\). Since the bearing of \(C\) from \(B\) is \(130^\circ\), the interior angle \(\angle ABC\) of the triangle is the difference between these bearings: \(\angle ABC = 240^\circ - 130^\circ = 110^\circ\). We now use the Cosine Rule in triangle \(ABC\) to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\). Substituting the given lengths: \(AC^2 = 15^2 + 22^2 - 2 \times 15 \times 22 \times \cos(110^\circ) = 225 + 484 - 660 \times (-0.34202) = 709 + 225.733 = 934.733\). Taking the square root: \(AC = \sqrt{934.733} \approx 30.573...\text{ km}\). Correct to 3 significant figures, the distance is \(30.6\text{ km}\).

M1: for finding the interior angle \(\angle ABC = 110^\circ\) (or showing a diagram with correct angle calculations). A1: for \(\angle ABC = 110^\circ\). M1: for substituting values into the Cosine Rule: \(AC^2 = 15^2 + 22^2 - 2 \times 15 \times 22 \times \cos(110^\circ)\). A1: for 30.6 (accept answers in range 30.5 - 30.6).

Let the amount of money invested in schemes \(P\), \(Q\) and \(R\) be represented by \(400\), \(500\) and \(300\) respectively.

1. **Calculate the total initial investment:**
\[\text{Total initial investment} = 400 + 500 + 300 = 1200\]

2. **Calculate the changes in value for each scheme:**
* For scheme \(P\): \(400 \times 0.08 = +32\)
* For scheme \(Q\): \(500 \times (-0.04) = -20\)
* For scheme \(R\): \(300 \times 0.15 = +45\)

3. **Calculate the net change in the total investment:**
\[\text{Net change} = +32 - 20 + 45 = +57\]

4. **Calculate the overall percentage increase:**
\[\text{Overall percentage increase} = \frac{57}{1200} \times 100 = 4.75\%\]

Alternatively, using algebraic representation with \(x\):
Let the investments be \(4x\), \(5x\) and \(3x\), giving a total of \(12x\).
* New value of \(P = 4x \times 1.08 = 4.32x\)
* New value of \(Q = 5x \times 0.96 = 4.80x\)
* New value of \(R = 3x \times 1.15 = 3.45x\)
* Total new value \(= 4.32x + 4.80x + 3.45x = 12.57x\)
* Percentage increase \(= \frac{12.57x - 12x}{12x} \times 100 = 4.75\%\)

**M1**: For a method to represent the investments using a common variable or specific values in the ratio \(4:5:3\) (e.g., \(4x, 5x, 3x\) or \(400, 500, 300\)), and stating the total initial investment as \(12x\) or \(1200\).

**M1**: For calculating the change or the new value for at least two of the schemes (e.g., \(400 \times 0.08 = 32\) or \(432\), and \(500 \times (-0.04) = -20\) or \(480\)).

**M1**: For a complete method to find the overall percentage increase, e.g., \(\frac{32 - 20 + 45}{1200} \times 100\) or \(\frac{1257 - 1200}{1200} \times 100\).

**A1**: For \(4.75\) (or \(4.75\%\), or equivalent fraction/mixed number like \(\frac{19}{4}\) or \(4\frac{3}{4}\)).